# Derivative of a fraction of two complex matrix production

Suppose $$\mathbf{a}_{N\times1}$$ is a complex-valued vector.

$$B_{K\times N}$$ is a complex-valued matrix.

$$C_{K\times K}$$ is a real-valued diagonal matrix.

Here is a function $$f: \mathbb{C}^N \to \mathbb{R}$$, $$f = \frac{\mathbf{a}^H B^H C B \mathbf{a}}{\mathbf{a}^H B^H B \mathbf{a}}$$, and whose numerator and denominator both are real-valued scalars (this can be verified). Here the operator $$H$$ means conjugate transpose.

Now I want to take the derivative of function $$f$$, but I only have the formula for real vectors like $$\frac{\partial (\mathbf{x}^T B \mathbf{x})}{\partial \mathbf{x}} = (B+B^T)\mathbf{x}$$, which seems not applicable in this case. How can I do this? And further, if I let the derivative of this function $$f$$ equal to zero, can I get the max or min value of the function $$f$$?

Thank you!

• So, you want to take the derivative of $f$ with respect to $a$? – user550103 Mar 11 at 16:56
• Yes you’re right! Could you help me? – J L Mar 11 at 16:58

HINT (well, it's nearly complete)

Some definitions with differentials.

$$D := B^H B$$ and $$N := B^H C B$$

$$\alpha := a^H N a = \operatorname{Tr}(a^H N a) := a^* : Na,$$ where notation $$:$$ corresponds to Frobenius product and the operator $$(\cdot)^*$$ is complex conjugate.

Differential of $$\alpha$$ is $$\ d \alpha = Na : da^*$$

Similarly, $$\beta := a^H D a$$ $$\Rightarrow d\beta = Da : da^*$$

$$\gamma:= \beta^{-1}$$ $$\Rightarrow d\gamma = -\beta^{-2} \ d\beta$$

Now, you can describe your function as $$f := \gamma \alpha \ .$$

Take the differential and then we obtain the gradient, i.e., \begin{align} df &= d\gamma \ \alpha + \gamma \ d \alpha\\ &= \left( -\beta^{-2} \ d\beta \right) \ \alpha + \gamma \ \left( Na : da^* \right)\\ &= -\beta^{-2} \alpha \ \left( Da : da^* \right) + \gamma \ \left( Na : da^* \right)\\ &= \left( -\beta^{-2} \alpha \ Da + \gamma Na \right): da^* \ . \end{align}

The gradient is \begin{align} \frac{\partial f}{\partial a^*} &= \left( -\beta^{-2} \alpha \ D + \gamma N \right)a \ . \end{align}

• Since $f$ is presumed real, the other gradient (if needed) is $$\frac{\partial f}{\partial a} = \left(\frac{\partial f}{\partial a^*}\right)^*$$ – greg Mar 11 at 19:18
• @greg: Thank you. I missed that one. :). – user550103 Mar 11 at 19:40
• @user550103 Thank you for your kind help! – J L Mar 12 at 3:21